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  • Alyona and towers CodeForces

    大意: 给定序列, 要求实现区间加, 询问整个序列最长的先增后减的区间.

    线段树维护左右两端递增,递减,先增后减的长度即可, 要注意严格递增, 合并时要注意相等的情况, 要注意相加会爆int.

    #include <iostream>
    #include <random>
    #include <algorithm>
    #include <cstdio>
    #include <math.h>
    #include <set>
    #include <map>
    #include <queue>
    #include <string>
    #include <string.h>
    #include <bitset>
    #define REP(i,a,n) for(int i=a;i<=n;++i)
    #define PER(i,a,n) for(int i=n;i>=a;--i)
    #define hr putchar(10)
    #define pb push_back
    #define lc (o<<1)
    #define rc (lc|1)
    #define mid ((l+r)>>1)
    #define ls lc,l,mid
    #define rs rc,mid+1,r
    #define x first
    #define y second
    #define io std::ios::sync_with_stdio(false)
    #define endl '
    '
    #define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
    using namespace std;
    typedef long long ll;
    typedef pair<int,int> pii;
    const int P = 1e9+7, INF = 0x3f3f3f3f;
    ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
    ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
    ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
    inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
    //head
    
    
    
    const int N = 3e5+10;
    int n, m;
    struct _ {
    	ll l,r,tag;
    	int len,La,Lb,Lab,Ra,Rb,Rab,ab;
    	_ () {}
    	_ (int x) {
    		l=r=x,len=La=Lb=Lab=Ra=Rb=Rab=ab=1,tag=0;
    	}
    	void upd(ll x) {
    		l+=x,r+=x,tag+=x;
    	}
    	_ operator + (const _ &rhs) const {
    		_ ret;
    		ret.l=l,ret.r=rhs.r;
    		ret.len=len+rhs.len;
    		ret.La = La+(La==len&&r<rhs.l?rhs.La:0);
    		ret.Lb = Lb+(Lb==len&&r>rhs.l?rhs.Lb:0);
    		ret.Lab = Lab;
    		if (La==len) { 
    			if (r<rhs.l) ret.Lab=La+rhs.Lab;
    			else if (r>rhs.l) ret.Lab=La+rhs.Lb;
    		}
    		else if (Lab==len&&r>rhs.l) ret.Lab=Lab+rhs.Lb;
    		ret.Ra = rhs.Ra+(rhs.Ra==rhs.len&&r<rhs.l?Ra:0);
    		ret.Rb = rhs.Rb+(rhs.Rb==rhs.len&&r>rhs.l?Rb:0);
    		ret.Rab = rhs.Rab;
    		if (rhs.Rb==rhs.len) {
    			if (r<rhs.l) ret.Rab=rhs.Rb+Ra;
    			else if (r>rhs.l) ret.Rab=rhs.Rb+Rab;
    		}
    		else if (rhs.Rab==rhs.len&&r<rhs.l) ret.Rab=rhs.Rab+Ra;
    		ret.ab = max(ab,rhs.ab);
    		if (r!=rhs.l) ret.ab=max(ret.ab,Ra+rhs.Lb);
    		if (r<rhs.l) ret.ab=max(ret.ab,Ra+rhs.Lab);
    		if (r>rhs.l) ret.ab=max(ret.ab,Rab+rhs.Lb);
    		ret.tag = 0;
    		return ret;
    	}
    } tr[N<<2];
    void build(int o, int l, int r) {
    	if (l==r) tr[o]=_(rd());
    	else build(ls),build(rs),tr[o]=tr[lc]+tr[rc];
    }
    void pd(int o) {
    	if (tr[o].tag) {
    		tr[lc].upd(tr[o].tag);
    		tr[rc].upd(tr[o].tag);
    		tr[o].tag=0;
    	}
    }
    void update(int o, int l, int r, int ql, int qr, int v) {
    	if (ql<=l&&r<=qr) return tr[o].upd(v);
    	pd(o);
    	if (mid>=ql) update(ls,ql,qr,v);
    	if (mid<qr) update(rs,ql,qr,v);
    	tr[o]=tr[lc]+tr[rc];
    }
    int main() {
    	build(1,1,n=rd());
    	m=rd();
    	REP(i,1,m) {
    		int l=rd(),r=rd(),d=rd();
    		update(1,1,n,l,r,d);	
    		printf("%d
    ", tr[1].ab);
    	}
    }
    
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  • 原文地址:https://www.cnblogs.com/uid001/p/10799452.html
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